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Along the way to a much simpler solution to a homology problem, I thought about computing the fundamental group of $S^2 / A$. I quickly ran into trouble, so I want to know if there is there a slick way to do this. (For nontrivial |A|, of course.)

I don't think that Van Kampen's theorem works, since the intersection will not be path connected for $|A| \geq 2$. (At least for the obvious covers.)

Elle Najt
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  • What are the «obvious» covers? – Mariano Suárez-Álvarez May 15 '14 at 19:36
  • For instance, partitioning the $X = S^2 / {x,y}$ around the reflective symmetry (seen by embedding $X$ into $R^3$ and choosing x and y the north and south poles) and enlarging those partitions to open covers. Then doing similarly for the rest. Or covering the bad point (where they are all glued together) with a contractible neighborhood, and enlarging the complement of that neighborhood to get a covering. – Elle Najt May 15 '14 at 19:40
  • Van Kampen works, just take neighborhoods of the points. If they're finite then you have a finite cover of S^2 so that intersection of every two neighborhoods is simply connected. Note that this doesn't work if you have infinite points cause $S^2 - {p} \simeq \mathbb{R}^2$ and $\mathbb{Q}^2 \subset \mathbb{R}^2$ is countable and dense (so you can't state that the intersection is simply connected). – Alessandro Flati May 15 '14 at 19:44
  • I think this is homotopy equivalent to a wedge of $S^2$ with $\left|A\right|-1$ copies of $S^1$. – Christoph May 15 '14 at 19:50
  • @AlessandroFlati I don't think their intersection will even be connected, unless you are choosing your neighborhoods very carefully as in Mariano's answer below. – Elle Najt May 15 '14 at 20:05
  • @Christoph How do you see that homotopy? – Elle Najt May 15 '14 at 20:10
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    @Christoph: If you remove $2$ points, it will be hommotopy equivalent to a circle, but not to a sphere wedged with a circle. – Stefan Hamcke May 15 '14 at 20:13
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    @StefanHamcke We are not removing points, but identifying via a quotient. Sorry, the set minus and set quotient symbols are a little ambiguous. – Elle Najt May 15 '14 at 20:18
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    Ouch, I didn't read carefully enough, sorry. Well, then @Christoph is right. – Stefan Hamcke May 15 '14 at 20:19
  • Oh, I can see the homotopy now. Thanks for pointing that out @Christoph. – Elle Najt May 15 '14 at 20:33
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    Identifying two points is homotopy equivalent to gluing a line segment that connects the points, and that's homtopy equivalent to wedging with a circle, as long as both points are in the same path connected component. – Christoph May 15 '14 at 20:48

2 Answers2

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Let us do first the following simple computation. Let $D$ be the closed unit disc in the plane and let $A$ be a finite subset of the interior of $D$. What is $\pi_1(D/A)$?

Let $n=|A|$. It does not matter which $n$ points are in $A$, only the cardinal of the set, so we may assume that the points of $A$ are the $n$ vertices of a regular polygon centerd at the origin. Let us take basepoint in $D/A$ to be the (image of) the origin. There is an obvious retract of the space $D/A$ to the space obtaned from a regular $n$-gon by identifing its vertices. This looks like a parachute.

This space can clearly be constructed as a CW-complex as follows: start with one vertex. Next add $n$ $1$-cells in the form of loops, and now glue a $2$-cell in the obvious way. The fundamental group of this space is then $\langle x_1,\dots,x_n:x_1\cdots x_n=1\rangle$. This is easily seen to be a free group in $n-1$ generators.

Now your sphere can be covered with two open sets. One of which is an open disc and the other an open set which deformation retracts to my $D/A$. The intersection is a (thick) circle. Now use van Kampen.

Mariano Suárez-Álvarez
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  • So the fundamental group is free on $n - 1$ generators, because the loop in the intersection is trivial in the space which deformation retracts to $D / A$? – Elle Najt May 15 '14 at 20:01
  • Incidentally, I found this much easier to visualize (up to homotopy) by connecting glued points with lines instead of trying to pinch them together. – Elle Najt May 15 '14 at 20:05
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It seems standard texts fail to take into account the algebraic modelling by the theory of groupoids.

There is a nice construction introduced by Philip Higgins, see his downloadable book Categories and Groupoids, Chapter 8, in which you start with a groupoid $G$ and a function $f: Ob(G) \to Y$ where $Y$ is any set. Then there is a groupoid which he writes $U_f(G)$ with object set $Y$ and with a universal property for morphisms $G \to H$ whose function on objects factors through $f$. Thus $U_f(G)$ is obtained from $G$ by "identifying certain objects of $G$", and perhaps adding some. This construction includes that of free groups, and of free products of groups.

The topological interpretation in terms of the fundamental groupoid $\pi_1(Z,C)$ of a space $Z$ for a set $C$ of base points is given on p.343, 9.1.2 (Corollary 3) of Topology and Groupoids, and was in the 1968 edition of this book.

This answers the special case of the question, as given by Mariano.

Later: I think also that my and other, answers to this mathoverflow question on many base points and groupoids are relevant.

Community
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Ronnie Brown
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  • Thanks for these resources - I'll make some time to read them. I have seen the fundamental groupoid developed theoretically, but I don't have any experience doing computations with it. I guess I should develop that. – Elle Najt May 17 '14 at 20:07
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    You might have a look at arXiv:1404.0556 for a recent use of groupoids. There are not many texts which mention free groupoids on (directed) graphs, an obvious enough idea in this day and age. Good luck! – Ronnie Brown May 17 '14 at 22:01